Simple? Statics 2

[Lcubed]

A load is imposed at the center of a square 4-bolt pattern. One bolt is removed. How is the load distributed to the remaining 3 bolts?

I know the location of the "new" centroid of the fastener pattern, and I have calculated the moments about that point and the resulting reactions. I have also looked at the problem using "rigid body" methods, per Maleev and others.

However, I think that the load lies on a line between two fasteners, and there is no moment about that line, and so those two fasteners take all of the load while the third remaining bolt takes no load at all. I need a reference to show that I am right or wrong, as the case may be. Can anyone help?

 

[desertfox]

Hi Lcubed

Look at this thread:- Thread2-101606.

I am assuming your load is at right angles to the bolt heads is that correct?

If my assumption is correct both you and I have contributed
threads to the link above.


Regards Desertfox

 

[Lcubed]

I'm saying that the load didn't move, thus the orientation should not matter. The load is equidistant from all three remaining bolts, and lies at the middle of a line between the two bolts which are farthest apart. I may not have been clear about this in my write-up. Does this help?
Regards,
Lcubed

 

[loquillo]

I can not visualize your problem. Is it vert or hor, how is it supported, is it a distributed load or a localized load.
There are many solutions to your problem depending on the arrangement of the bolt/plate system. I think that most of the load can be ditributed between 2 bolts but the third bolt also carries load even though is negligible in comparison with the load carried by the other 2 bolts.

 

[desertfox]

Hi Lcubed

Orientation of the load is important; ie if the load is at right angles to the bolt axis you have shear across thr bolt diameters, if the load is parallel to axis of the threads you have tension and a prying action.

regards desertfox

 

[Lcubed]

Hi desertfox,
Right. Glad to hear from you. The problem that I am having trouble with is that this load, on the line between the two most widely separated remaining bolts, does not cause a moment about the line, and therefore doesn't seem likely to induce a reaction in the third remaining bolt. Yet, using the approach I recommended in the thread you cited, one gets 1/3 of the load on each bolt because the bolts are equidistant from the load. Being a little more realistic, and considering the load eccentric from the centroid of the pattern, one gets a different answer. But I still don't see how there can be any load at all on that third bolt???
Regards,
Lcubed

 

[Lcubed]

Hello, All,
I should have been more clear. Think of this as a flat horizontal plate hanging on four nuts attached to vertical tension rods extending from the ceiling. Say that the rods are at the corners of a square pattern, one foot on each side. It is 1.414 feet from a rod to the rod diagonally opposite. A load hangs from the center of the plate; it is .707 foot from each rod. There is no shear. The load is parallel to the axes of the rods. There is no prying. Consider this a rigid body problem with no deformation-just pure statics. Hope this helps.
Regards,
Lcubed

 

[EnglishMuffin]

Assuming that all the load would be carried on only two of the three bolts is certainly conservative, but if your question is what the loads would actually be in practice, you would have to take into account the stiffness of the two components you are bolting together, and the bolt preload. In many cases, with properly designed and preloaded joints, the applied load often has very little effect on the bolt stresses.

 

[Lcubed]

Further clarification: Change the bolts to cables.
Regards,
Lcubed

 

[profengmen]

I did a quick FEA on a model, a simple plate with three holes and a load point directly in the middle...it's all I really had to go on.

As I suspected, there is stress on the third bolt directly perpendicular to the load. The deformation of the plate looks exactly as I suspected it would, the unsupported corner had the largest deflection.

The material is going to deform, and when it does, stress is going to occur at all restraining points.

The two holes opposite each other had stresses spread out in three different directions.

 

[zekeman]

Profengman,
There is no stress in the 3rd bolt if it is suspended by a cable (tensile only) since as we all know the moments about the line of centers of the two inline bolts must be zero.
Your model assumed a restraint at that third bolt which would depend on the actual structure of the third supporting member(a rod or a cable).

 

[profengmen]

zekeman,
I didn't read the "no deformation" part of the thread... My model assumed deformation, which was stated in my thread.

Then again, that's an unrealistic situation...

 

[btrueblood]

Okay, let's use cables. Four cables, supporting a tensile load on a suspended plate. One cable breaks. The off-axis cable still had tension from the original load, so it "pulls" sideways until the moment of the plate balances its tension. Same answer as profengmen gave: in the real world, all of the connections absorb some of the load, and all of the connections are elastic, and will respond to changes in loading/pattern.

Ben T

 

[zekeman]

Ben,
Reread my thread. If there is only tensile support, the force in the offline bolt is zero. It doesn't care about the previous history of the suppot system.

 

[brad]

I don't think there is enough information to definitely determine the answer--it is a statically indeterminate system (thus it's not "pure statics"). The answer is dependent on the stiffness of the piece to which the bolts are attached.

If all bolts are attached to an effectively rigid piece, then I think I would agree that only two bolts take the force. However, we don't know whether rigid is a reasonable assumption.
Brad

 

[Dennyd]

Hooke's Law says these plates will not be "perfectly rigid". Statically indeterminate - yes. Solvable - yes.

It sounds like this might be purely an academic question to satisfy someone's curiosity. Either you've got the appropriate engineering background to solve this or you don't. Since the plate has some deflection all three bolts will have some loading.

- - -Dennyd, P.E.

 

[btrueblood]

Zekeman said:

"Reread my thread. If there is only tensile support, the force in the offline bolt is zero. It doesn't care about the previous history of the suppot system."

I have read and re-read your posts. Please note that what at least three people here are telling you (multiple times, and two of them purport to be PE's) is not BS. Real world members have elasticity (they stretch).

If, before applying a load, all four cables are taut, then each cable MUST STRETCH when the load is applied. This stretch develops a certain amount of tensile force in each cable. When one cable is then severed or loosened, the remaining cables react to the change in load pattern. I can not envision any combination of cable patterns or distribution of tensile forces where you can say that the third cable will have zero force -- its residual tension reacts to removal of the opposite force, pulling the plate "out of plumb" and producing a moment (due to the change in relative position of the downwards pulling force).

If you are now going to say that two cables have slack before the load is applied (and thus have zero tension at the loaded condition), then you have changed the problem.

I suggest you set this problem up with some strings and a block of wood, and some weights. Let us all know the results of your experiment.

 

[zekeman]

You don't need a PE license or any other license to know that if we are talking cables, the only forces are tensile in the cable and, again, there can be no force in the offline cable, since, if there were, you would get a moment about the line of centers, thus violating the equilibrium condition.

ssk, PE

 

[zekeman]

Oh, I forgot to mention,did it ever occur to you that after the cable above the 4th bolt snaps, the plate will rotate about the line of centers of the load and the inline bolts, removing the load from the bolt in question. So if the plate has no weight, I stick to my conclusion. If you include the weight of the plate, then, yes, the forces would be distributed, according to the laws of statics.

 

[btrueblood]

Fine, you win. Go build it.